Question

please solve the sum...​

Answer 1

Answer:

$\longrightarrow\sf{ \dfrac{3}{x - 1} + \dfrac{1}{x - 3} = \dfrac{4}{x - 2}}$

$\\ \longrightarrow\sf{ \dfrac{3(x - 3) + 1(x - 1)}{(x - 1) (x - 3)} = \dfrac{4}{x - 2}}$

$\\ \longrightarrow\sf{ \dfrac{3x - 9 + x - 1}{x(x - 3) - 1(x - 3)} = \dfrac{4}{x - 2}}$

$\\ \longrightarrow\sf{ \dfrac{3x + x - 9 - 1}{{x}^{2} - 3x - x + 3} = \dfrac{4}{x - 2}}$

$\\ \longrightarrow\sf{ \dfrac{4x - 10}{{x}^{2} - 4x + 3} = \dfrac{4}{x - 2}}$

$\\ \longrightarrow\sf{ (4x - 10) (x - 2) = 4({x}^{2} - 4x + 3)}$

$\\ \longrightarrow\sf{4x ( x -2 ) - 10 (x - 2) = 4{x}^{2} - 16x + 12}$

$\\ \longrightarrow\sf{4{x}^{2} - 8x - 10x + 20 = 4{x}^{2} - 16x + 12}$

4x² get cancelled on both sides .

$\\ \longrightarrow\sf{ - 18x + 20 = - 16x + 12}$

$\\ \longrightarrow\sf{ - 18x + 16x = 12 - 20}$

$\\ \longrightarrow\sf{ -2x = - 8}$

$\\ \longrightarrow\sf{ x = \dfrac{- 8}{-2}}$

$\\ \longrightarrow\sf{ x = 4}$